## Altitude

In triangle $ABC, BC=13, CA=14$, and $AB=15$. If $D$ is a point on $\overline{CA}$ such that $\overline{BD}$ is perpendicular to $\overline{CA}$, what is $BD$?

Source: NCTM Mathematics Teacher, November 2006

SOLUTION
Area of triangle $ABC$ by Heron’s formula
$s=(1/2)(13+14+15)=21$
Area of triangle $ABC=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{7056}$
$=84$
Area of triangle $ABC$ using base $AC$ and altitude $BD$
$(1/2)14\times BD=84$
$BD=12$

Answer: $12$

Alternative solution 1

Use the Pythagorean theorem in right triangle $CDB$
$x^2+h^2=13^2=169\qquad\qquad (1)$
Similarly, in right triangle $ADB$
$(14-x)^2+h^2=15^2$
$196-28x+x^2+h^2=225$
$x^2+h^2=29+28x$
From Eq. $(1)$
$169=29+28x$
$x=5$
Substitute the value of $x$ into Eq. $(1)$
$5^2+h^2=169$
$h^2=144$
$h=12$

Alternative solution 2

Use the Pythagorean theorem in right triangles $ADB$ and $CDB$
$DA^2+BD^2=15^2=225\qquad\qquad (2)$
$CD^2+BD^2=13^2=169\qquad\qquad (3)$
Subtract Eq. $(3)$ from Eq. $(2)$
$DA^2-CD^2=56$
$(DA+CD)(DA-CD)=56$
$AC(DA-CD)=56$
$14(DA-CD)=56$
$DA-CD=4\qquad\qquad\;\;(4)$
$DA+CD=14\qquad\qquad (5)$
Add Eq. $(4)$ and Eq. $(5)$
$2DA=18$
$DA=9$
Substitute the value of $DA$ into Eq. $(2)$
$9^2+BD^2=225$
$BD^2=144$
$BD=12$

Alternative solution 3

Use the law of Cosine in triangle $ABC$
$15^2=13^2+14^2-2(13)(14)\text{cos}\,C$
$225=169+196-2(182)\text{cos}\,C$
$225=365-364\,\text{cos}\,C$
$\textrm{-}140=\textrm{-}364\,\text{cos}\,C$
$\text{cos}\,C=140/364$
In right triangle $CDB$
$\text{cos}\,C=CD/BC=CD/13$
$140/364=CD/13$
$CD=5$
Use the Pythagorean theorem in right triangle $CDB$
$CD^2+BD^2=BC^2$
$5^2+BD^2=13^2$
$BD^2=144$
$BD=12$